Problem: Sector $OAB$ is a quarter of a circle of radius 3 cm. A circle is drawn inside this sector, tangent at three points as shown. What is the number of centimeters in the radius of the inscribed circle? Express your answer in simplest radical form. [asy]
import olympiad; import geometry; size(100); defaultpen(linewidth(0.8));
draw(Arc(origin,3,90,180));
draw((-3,0)--(origin)--(0,3));
real x = 3/(1 + sqrt(2));
draw(Circle((-x,x),x)); label("$B$",(0,3),N); label("$A$",(-3,0),W);
label("$O$",(0,0),SE); label("3 cm",(0,0)--(-3,0),S);
[/asy]
Solution: Call the center of the inscribed circle $C$, and let $D$ be the point shared by arc $AB$ and the inscribed circle.  Let $E$ and $F$ be the points where the inscribed circle is tangent to $OA$ and $OB$ respectively.  Since angles $CEO$, $CFO$, and $EOF$ are all right angles, angle $FCE$ is a right angle as well.  Therefore, the measure of angle $DCE$ is $(360-90)/2=135$ degrees.  By symmetry, angles $ECO$ and $FCO$ are congruent, so each measures 45 degrees.  Therefore, angle $DCO$ measures $135+45=180$ degrees, which implies $DC+CO=OD$.  Also, $DC=r$, and $CO=r\sqrt{2}$, since triangle $CEO$ is an isosceles right triangle. Since $OD$ is a radius of the circle centered at $O$, we may set $DC+CO=r+r\sqrt{2}$ equal to 3 cm to find \[
r=\frac{3\text{ cm}}{\sqrt{2}+1}\cdot\frac{\sqrt{2}-1}{\sqrt{2}-1}=\boxed{3\sqrt{2}-3}\text{ centimeters}.
\]

[asy]
import olympiad; import geometry; size(150); defaultpen(linewidth(0.8));
draw(Arc(origin,3,90,180));
draw((-3,0)--(origin)--(0,3));
real x = 3/(1 + sqrt(2));
draw(Circle((-x,x),x)); label("$B$",(0,3),N); label("$A$",(-3,0),W);
label("$O$",(0,0),SE); draw((0,0)--(-3,0));
label("$C$",(-x,x),NE); label("$D$",(-3/sqrt(2),3/sqrt(2)),NW);
label("$F$",(0,x),E); label("$E$",(-x,0),S);
draw((-x,0)--(-x,x)--(0,x));
draw((-x,x)--(-3/sqrt(2),3/sqrt(2)));
draw((-x,x)--origin,linetype("1 2"));[/asy]